Question

高一数学 详细写下过程

设a,b∈(0,+∞),a+b=1,求证:√(a+1/2)+√(b+1/2)≤2

Answer 1

a+1/2+b+1/2=2,
[(a+1/2)^(1/2)]*[(b+1/2)^(1/2)]<=[a+1/2加上b+1/2]/2=1,
f^2={(a+1/2)^(1/2)+(b+1/2)^(1/2)}^2=a+1/2加上b+1/2+2*[(a+1/2)^(1/2)]*[(b+1/2)^(1/2)]=2+2*[(a+1/2)^(1/2)]*[(b+1/2)^(1/2)]<=2+2*1=4=2^2,f=(a+1/2)^(1/2)+(b+1/2)^(1/2)<=2,此即：根号下a+1/2加上根号下b+1/2<=2
PS：[(a+1/2)^(1/2)]即根号下(a+1/2)

Answer 2

√(a+1/2)+√(b+1/2)≤2√[(a+1/2+b+1/2)/2]=2
当且仅当a+1/2=b+1/2 即a=b=1/2时等号成立
【其中用到(m+n)/2≤√[(m^2+n^2)/2]】